BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
In ΔBEC and ΔCFB,
∠BEC = ∠CFB (Each 90°)
BC = CB (Common)
BE = CF (Given)
∴ ΔBEC ≅ ΔCFB (By RHS congruency)
⇒ ∠BCE = ∠CBF (By CPCT)
∴ AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, ΔABC is isosceles.
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